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速度规划问题 - 扩展模型

在上述例子中, 我们考虑了简单的速度规划问题的建模与求解. 下面我们将模型扩展为一个更复杂的模型:

  • 考虑控制量的变化率约束/惩罚
  • 考虑变量的耦合约束
  • l1范数的目标函数

变化率约束/惩罚​

假设控制量aa的变化率的约束为[-5, 5] m/s3m/s^3. 同时, 我们需要在目标函数中添加控制量变化率的惩罚项.

我们可以引入控制量的高阶导数变量jerkjerk来表示控制量的变化率. 然后对于jerkjerk变量, 我们设置其上下界为[-5, 5] m/s3m/s^3, 并且对其进行惩罚.

新增变量:

jerk = prob.variable('jerk', hard_lowerbound=-5, hard_upperbound=5)

目标函数变更为:

obj = general_objective((v - vref)**2 + 0.1 * jerk**2)
prob.objective(obj)

约束条件变更为:

ode = differential_equation(
    state=[s, v, a],
    state_dot=[v, a, jerk], 
    stepsize=0.1, 
    discretization_method='erk4')
prob.equality(ode)

初始条件变更为 (增加了加速度的初始值约束):

seq = general_equality([s - 0, v - 0, a - 0])
prob.start_equality(seq)

效果如下:

speed_jerk

耦合约束​

假设我们需要在[7, 8] s的时间窗口内, 希望距离约束保留一定的随速度变化的裕度:

s(t)≥60+0.2v(t),t∈[7,8] \begin{split} &s(t) \geq 60 + \textcolor{red}{0.2 v(t)}, t \in [7, 8] \\ \end{split}

这里我们使用了general_inequality函数来定义耦合约束.

tsafe = prob.parameter('tsafe',  stage_dependent=True)
ineq = general_inequality([s - tsafe * v - smin], sign=['>='], bound=[0.0])
prob.inequality(ineq)

同样, 在求解时, 我们需要设置[7, 8] s间的tsafe的值为0.2.

for (size_t i = 0; i < SPEED_PLANNING_DIM_N; i++) {
    /* smin, tsafe */
    if ((i >= 70) && (i <= 80)) {
        prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_SMIN] = 60.0;
        prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_TSAFE] = 0.2;
    } else {
        prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_SMIN] = 0.0;
        prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_TSAFE] = 0.0;
    }
}

效果如下:

speed_sv_ineq

l1范数的目标函数​

假设我们需要将目标函数变更为l1范数的形式:

min⁡∑i=1N∣vi−vref∣+0.1∣jerki∣ \min \sum_{i=1}^{N} |v_i - v_{ref}| + 0.1 |jerk_i|

注意, 上述目标函数不是一个smooth的目标函数, 我们需要对其进行等效转化. 通过引入slack变量uvu_v与ujerku_{jerk}, 上述目标函数可以转化为:

min⁡∑i=1Nuv+0.1ujerk \min \sum_{i=1}^{N} u_v + 0.1 u_{jerk}

同时, 我们需要添加以下的约束条件:

uv≥v−vrefuv≥−(v−vref)ujerk≥jerkujerk≥−jerk \begin{split} &u_v \geq v - v_{ref} \\ &u_v \geq - (v - v_{ref}) \\ &u_{jerk} \geq jerk \\ &u_{jerk} \geq - jerk \\ \end{split}

对于上述优化问题, 我们可以使用OPTIMake内置的软约束接口等效实现. 这样我们就不需要显示地定义slack变量, 实现内存占用与计算时间的减少.

对于jerkjerk变量, 上述可以等效于一个上下限都为0的l1软约束:

jerk = prob.variable('jerk', hard_lowerbound=-5, hard_upperbound=5, 
                     soft_lowerbound=0, soft_upperbound=0, 
                     weight_soft_lowerbound=0.1, weight_soft_upperbound=0.1,
                     penalty_type_soft_lowerbound='l1',
                     penalty_type_soft_upperbound='l1')

对于vv变量, 上述也可以等效于一个上下限都为0的l1软约束:

ineq = general_inequality([s - tsafe * v - smin, v - vref, v - vref], 
                           sign=['>=', '>=', '<='], 
                           bound=0.0)
prob.inequality(ineq, penalty_type=['none', 'l1', 'l1'], weight_soft=[inf, 1.0, 1.0])

效果如下, 可以看到l1范数的目标函数与二次型的目标函数相比, 可以使得加减速更加线性.

speed_l1

附录​

以上建模的完整代码如下:

prob = multi_stage_problem('speed_planning', 100)
vref = prob.parameter('vref',  stage_dependent=False)
smin = prob.parameter('smin',  stage_dependent=True)
tsafe = prob.parameter('tsafe',  stage_dependent=True)

s = prob.variable('s', hard_lowerbound=smin)
v = prob.variable('v')
a = prob.variable('a', hard_lowerbound=-3, hard_upperbound=3)
jerk = prob.variable('jerk', hard_lowerbound=-5, hard_upperbound=5, 
                     soft_lowerbound=0, soft_upperbound=0, 
                     weight_soft_lowerbound=0.1, weight_soft_upperbound=0.1,
                     penalty_type_soft_lowerbound='l1',
                     penalty_type_soft_upperbound='l1')


# use differential equation to define the state equation
ode = differential_equation(
    state=[s, v, a],
    state_dot=[v, a, jerk], 
    stepsize=0.1, 
    discretization_method='erk4')
prob.equality(ode)

# alternative way to define the state equation
# h = 0.1
# dis_eq = discrete_equation(
#     expr_next_stage=[s, v, a],
#     expr_this_stage=[s + h * v + 0.5 * h**2 * a + 1/6 * h**3 * jerk,
#                      v + h * a + 0.5 * h**2 * jerk,
#                      a + h * jerk],
# )
# prob.equality(dis_eq)

ineq = general_inequality([s - tsafe * v - smin,
                           v - vref, v - vref], 
                           sign=['>=', '>=', '<='], bound=0.0)
prob.inequality(ineq, penalty_type=['none', 'l1', 'l1'], weight_soft=[inf, 1.0, 1.0])

seq = general_equality([s - 0, v - 0, a - 0])
prob.start_equality(seq)

option = codegen_option()
codegen = code_generator()
codegen.codegen(prob, option)

以上求解的完整代码如下:

#include "speed_planning_prob.h"
#include "speed_planning_solver.h"
#include <stdio.h>

int main(void)
{
    Speed_planning_Problem prob;
    Speed_planning_Option option;
    Speed_planning_WorkSpace ws;
    Speed_planning_Output output;
    speed_planning_init(&prob, &option, &ws);

    /* params */
    prob.param[SPEED_PLANNING_PARAM_VREF] = 10.0; /* vref */
    for (size_t i = 0; i < SPEED_PLANNING_DIM_N; i++) {
        /* smin, tsafe */
        if ((i >= 70) && (i <= 80)) {
            prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_SMIN] = 60.0;
            prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_TSAFE] = 0.2;
        } else {
            prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_SMIN] = 0.0;
            prob.param_stage[i][SPEED_PLANNING_PARAM_STAGE_TSAFE] = 0.0;
        }
    }
    
    /* option */
    int solve_status = speed_planning_solve(&prob, &option, &ws, &output);
    
    printf("solve_status = %d\n", solve_status);
    return 0;
}